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Let $(\Omega, B, \mu)$ be a measure space. How can we characterize the space $(\Omega, B, \mu)$ so that

  1. the counting measure on $B$ is absolutely continuous with respect to $\mu$?
  2. the Dirac measure on $B$ is absolutely continuous with respect to $\mu$?

Edit: What I mean by the Dirac measure on $B$ is the following: Let $x\in \Omega$. Then the Dirac measure at $x$ assigns $1$ to a set in $B$ that contains $x$ and $0$ to a set that does not contain $x$.

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When you say "the dirac measure on $B$" what exactly do you mean? – Asaf Karagila Apr 12 '12 at 22:17
@AsafKaragila Isn't the dirac measure defined with respect to a point $x$? $$ \delta_x (A) := \begin{cases} 1 & x \in A \\ 0 & x \notin A \end{cases}$$ – Rudy the Reindeer Apr 12 '12 at 22:19
@MattN. That would be a Dirac measure, not the Dirac measure. – Asaf Karagila Apr 12 '12 at 22:20
I suppose @Michael Greinecker's answer is what you had in mind, but you should word you question more precisely because "What should $\Omega$, $B$, and $\mu$ be..." is too vague. That could mean that you just want any example, in which case you could take $\Omega$ to be any set, $B$ the full power set of $\Omega$, and in 1. $\mu$ the counting measure, and in 2. $\mu$ the Dirac measure concentrated at any $x \in \Omega$. The point is that every measure is absolutely continuous with respect to itself. – Quinn Culver Apr 12 '12 at 22:59
Okay, so this is the Dirac measure at $x$, not "the Dirac measure on $B$", the $x$ matters! :-) – Asaf Karagila Apr 12 '12 at 23:24

1 Answer 1

up vote 1 down vote accepted
  1. Counting measure is usually defined on the $\sigma$-algebra of all subsets. So let $\mu$ be another measure defined on all subsets. Counting measure is absolutely continuous with respect to $\mu$ if no $\mu$-zero set has positive counting measure. Every nonempty set has positive counting measure. So $\mu$ needs to put positve measure on every nonempty-set, in particular on every singleton. The measure on all singletons determines the measure on all countable sets and since uncountable sums of positve real numbers are always infinite, every such measure is determined by having a positive value at each singleton. So for $\mu$ there exists a function $f:\Omega\to\mathbb{R}\cup\{\infty\}$ with strictly positive values such that $\mu(A)=\sum_{\omega\in A}f(\omega)$ and this property characterizes the measures you are looking for.

  2. Dirac measures are bit more delicate since they can be defined on any $\sigma$-algebra. $\delta_x$ is absolutely continuous with respect to $\mu$ if $\mu(A)>0$ for every measurable set containing $x$. If the $\sigma$-algebra contains a smallest set $S$ containing $x$, which is true if it is countably generated or the powerset, then we just need $\mu(S)>0$.

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I'm a bit confused. what does $\mu$-zero set mean? – Josef Apr 12 '12 at 23:12
A measurable set $S$ with $\mu(S)=0$. – Michael Greinecker Apr 12 '12 at 23:14
Oh ok. Thanks. one more question. Could you please explain why $\delta_x \ll \mu$ if $\mu(A) \gt 0$ for every measurable set containing $x$? – Josef Apr 12 '12 at 23:20
Because these are exactly the ones that have positive measure under the dirac measure at $x$. – Michael Greinecker Apr 12 '12 at 23:29
Great. Thanks again. – Josef Apr 12 '12 at 23:34

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